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AP Physics 1 Unit 7 Review: Oscillations

Review AP Physics 1 Unit 7 to build a complete picture of oscillating systems, from the restoring force that defines simple harmonic motion to the energy exchanges that keep a spring or pendulum moving. This unit connects force, kinematics, and energy conservation in one unified framework.

Use the topic guides, practice questions, and FRQ practice available for this unit to work through SHM problems from multiple angles.

What is AP Physics 1 unit 7?

Oscillations appear whenever a restoring force pulls an object back toward equilibrium. Unit 7 formalizes that idea into simple harmonic motion, a model that applies to springs, pendulums, and many real systems. The unit builds from defining SHM, to calculating how fast a system oscillates, to representing the motion graphically, to tracking energy throughout the cycle.

Simple harmonic motion occurs when a restoring force is proportional to displacement from equilibrium. The key equations are m*a = -k*delta_x for the force condition, T = 2*pi*sqrt(m/k) for a spring, T = 2*pi*sqrt(L/g) for a pendulum, and E_total = (1/2)*k*A^2 for total mechanical energy. Kinetic energy is maximum at equilibrium; potential energy is maximum at the turning points.

What makes motion simple harmonic

SHM requires a linear restoring force: the net force on the object must point back toward equilibrium and be proportional to displacement. The equation m*a_x = -k*delta_x captures this. A pendulum with small angular displacement qualifies because the restoring torque is proportional to the angular displacement.

Period depends on system properties, not amplitude

For a spring-mass system, T = 2*pi*sqrt(m/k). For a simple pendulum, T = 2*pi*sqrt(L/g). Neither formula includes amplitude, which means doubling the amplitude of oscillation does not change how long each cycle takes. This is a frequently tested and counterintuitive result.

Energy shifts between kinetic and potential

Total mechanical energy in SHM is constant and equals (1/2)*k*A^2 for a spring system. At the turning points (x = plus or minus A), all energy is potential and kinetic energy is zero. At equilibrium (x = 0), all energy is kinetic and speed is maximum. Increasing amplitude increases total energy by a factor of A squared.

One framework, two systems

The same SHM model describes both spring-mass oscillators and small-angle pendulums. In both cases a restoring force or torque is proportional to displacement, the period is amplitude-independent, and energy continuously converts between kinetic and potential forms while the total stays constant. Recognizing this shared structure lets you transfer reasoning from one system to the other on any problem.

AP Physics 1 unit 7 topics

7.1

Defining Simple Harmonic Motion

SHM is defined by a linear restoring force: m*a_x = -k*delta_x. The force and acceleration are always directed toward equilibrium and are proportional to displacement. Pendulums with small angular displacement qualify under the same logic.

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7.2

Frequency and Period of SHM

T = 1/f. For a spring, T = 2*pi*sqrt(m/k). For a pendulum, T = 2*pi*sqrt(L/g). Amplitude does not appear in either formula, so changing amplitude does not change how long each oscillation takes.

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7.3

Representing and Analyzing SHM

Displacement, velocity, and acceleration all vary sinusoidally with the same period but different phases. Velocity is zero at turning points and maximum at equilibrium. Acceleration is always opposite to displacement. Amplitude changes the vertical scale of graphs but not the period.

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7.4

Energy of Simple Harmonic Oscillators

Total mechanical energy is constant at (1/2)*k*A^2. Energy shifts between kinetic and potential as the object oscillates. Kinetic energy peaks at equilibrium; potential energy peaks at the turning points. Total energy scales with the square of amplitude.

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practice snapshot

Hardest AP Physics 1 unit 7 topics

This snapshot uses Fiveable practice activity to show where students tend to miss questions and which review moves are worth prioritizing first.

62%average MCQ accuracy

Across 3.4k multiple-choice practice attempts for this unit.

3.4kMCQ attempts

Practice activity included in this snapshot.

41%average FRQ score

Across 10 scored free-response attempts for this unit.

Hardest topics in unit 7

MCQ miss rate
7.1

Review Defining Simple Harmonic Motion with attention to how the concept appears in AP-style source and evidence questions.

44%903 tries
7.2

Review Frequency and Period of SHM with attention to how the concept appears in AP-style source and evidence questions.

35%758 tries
7.4

Review Energy of Simple Harmonic Oscillators with attention to how the concept appears in AP-style source and evidence questions.

31%912 tries

Unit 7 review notes

7.1

The restoring force condition for SHM

SHM is a special case of periodic motion defined by its force law. The net force on the object must be a restoring force: it points opposite to the displacement from equilibrium and its magnitude is proportional to that displacement. The governing equation is m*a_x = -k*delta_x, where k is the spring constant and delta_x is displacement from equilibrium. Because acceleration is proportional to displacement and oppositely directed, the object accelerates most strongly at the turning points and has zero acceleration at equilibrium. A simple pendulum with small angular displacement fits this model because the restoring torque is proportional to the angular displacement, making the angular acceleration proportional to the angle.

  • Restoring force: A force directed opposite to the object's displacement from equilibrium; its magnitude increases with displacement, pulling the object back.
  • Equilibrium position: The location where net force on the object is zero; the object passes through this point with maximum speed.
  • m*a_x = -k*delta_x: The defining SHM force equation; the negative sign confirms the force opposes displacement, and k sets how stiff the restoring force is.
  • Small-angle approximation: For a pendulum, sin(theta) is approximately equal to theta in radians when the angle is small, making the restoring torque proportional to angular displacement and allowing the SHM model to apply.
A block on a spring is pulled 0.10 m from equilibrium and released. At what position is the net force on the block greatest in magnitude, and at what position is it zero?
SystemRestoring quantityProportionality condition
Spring-massForce F = -k*delta_xForce proportional to linear displacement
Simple pendulum (small angle)Torque tau proportional to thetaTorque proportional to angular displacement
7.2

Calculating period and frequency for springs and pendulums

Period T and frequency f are inverses: T = 1/f. For a spring-mass system, T_s = 2*pi*sqrt(m/k). For a simple pendulum with small angular displacement, T_p = 2*pi*sqrt(L/g). Neither formula contains amplitude, confirming that amplitude does not affect how long each oscillation takes. To predict how a change in the system affects the period, isolate the variable that changed: doubling mass increases T_s by a factor of sqrt(2); quadrupling spring constant cuts T_s in half; doubling pendulum length increases T_p by sqrt(2); changing amplitude has no effect on either period.

  • T = 1/f: Period and frequency are reciprocals; if a spring completes 2 oscillations per second, its period is 0.5 s.
  • T_s = 2*pi*sqrt(m/k): Period of a spring-mass oscillator; increases with more mass and decreases with a stiffer spring.
  • T_p = 2*pi*sqrt(L/g): Period of a simple pendulum; depends only on length and gravitational acceleration, not on the mass of the bob.
  • Amplitude independence: For both the spring-mass system and the small-angle pendulum, the period is the same regardless of how large or small the oscillation is.
A pendulum has period T on Earth. If you take it to a planet where g is four times larger, what is the new period in terms of T?
SystemPeriod formulaVariables that change TVariables that do NOT change T
Spring-mass2*pi*sqrt(m/k)Mass m, spring constant kAmplitude A
Simple pendulum2*pi*sqrt(L/g)Length L, gravitational acceleration gBob mass, amplitude A
7.3

Displacement, velocity, and acceleration graphs in SHM

Displacement in SHM follows x = A*cos(2*pi*f*t) or x = A*sin(2*pi*f*t) depending on initial conditions. Velocity and acceleration are also sinusoidal but shifted in phase. Velocity is zero at the turning points (x = plus or minus A) and maximum at equilibrium (x = 0). Acceleration is maximum in magnitude at the turning points and zero at equilibrium, always pointing opposite to displacement. On a graph, the x-t, v-t, and a-t curves all have the same period but peak at different times. Changing amplitude stretches the curves vertically but does not change the period or the horizontal spacing of the peaks and zeros.

  • x = A*cos(2*pi*f*t): Displacement equation for an oscillator starting at maximum positive displacement; A is amplitude and f is frequency.
  • Turning points: Positions x = plus or minus A where velocity is zero and acceleration (and restoring force) is maximum in magnitude.
  • Equilibrium crossing: Position x = 0 where speed is maximum and acceleration is zero; the object passes through fastest here.
  • Phase relationship: Velocity peaks one quarter cycle after displacement crosses zero; acceleration is always opposite in sign to displacement.
  • Amplitude vs. period: Increasing amplitude raises the maximum values of displacement, velocity, and acceleration but leaves the period unchanged.
An object in SHM has displacement x = 0.05*cos(4*pi*t) meters. At t = 0, what are the displacement, velocity direction, and acceleration direction?
QuantityAt x = +A (turning point)At x = 0 (equilibrium)
DisplacementMaximum positiveZero
SpeedZeroMaximum
Acceleration magnitudeMaximumZero
Restoring force magnitudeMaximumZero
7.4

Energy conservation in oscillating systems

The total mechanical energy of an SHM system is constant: E_total = U + K. For a spring-mass system, E_total = (1/2)*k*A^2. At the turning points, all energy is stored as spring potential energy U = (1/2)*k*x^2 and kinetic energy is zero. At equilibrium, all energy is kinetic and speed is at its maximum, v_max = A*sqrt(k/m). Because E_total scales with A squared, doubling the amplitude quadruples the total energy. This energy framework lets you find the speed of the object at any position using (1/2)*k*A^2 = (1/2)*m*v^2 + (1/2)*k*x^2.

  • E_total = (1/2)*k*A^2: Total mechanical energy of a spring-mass oscillator; set by the amplitude and spring constant, not by position or time.
  • U = (1/2)*k*x^2: Elastic potential energy at displacement x; maximum at the turning points and zero at equilibrium.
  • K = (1/2)*m*v^2: Kinetic energy; maximum at equilibrium and zero at the turning points.
  • v_max = A*sqrt(k/m): Maximum speed, reached when the object passes through equilibrium; increases with amplitude and decreases with more mass.
  • Energy and amplitude: Total energy is proportional to A squared, so a larger amplitude means more total energy stored in the system.
A spring with k = 200 N/m oscillates with amplitude 0.04 m. What is the total mechanical energy, and what is the speed of the mass at x = 0.02 m if the mass is 0.5 kg?
PositionKinetic energyPotential energyTotal energy
x = A (turning point)ZeroMaximum = (1/2)*k*A^2(1/2)*k*A^2
x = 0 (equilibrium)Maximum = (1/2)*k*A^2Zero(1/2)*k*A^2
x between 0 and APartialPartial(1/2)*k*A^2

Practice AP Physics 1 unit 7 questions

Try stimulus-based AP practice questions and written prompts after you review the notes.

Example stimulus-based MCQs

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Stimulus-based practice question

visual_answers

A simple pendulum oscillates with a small amplitude. The graph shows the horizontal velocity vv of the pendulum bob as a function of time tt.

Question

Which of the following graphs could represent the horizontal acceleration aa of the pendulum bob as a function of time tt?

Answer choice A
Answer choice B
Answer choice C
Answer choice D
visual_answers

Stimulus-based practice question

visual_answers

A block attached to a horizontal spring oscillates on a frictionless surface. The graph shows the block's position xx as a function of time tt.

Question

Which of the following graphs could represent the kinetic energy KK of the block as a function of time tt?

Answer choice A
Answer choice B
Answer choice C
Answer choice D

Example FRQs

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FRQ

Spring-mass oscillation energy conservation

2. A cart of mass 0.50 kg (see Figure 1) is attached to a spring and oscillates on a horizontal, frictionless track.

Figure 1. Cart–spring system at three specific positions: +0.20 m (released from rest), 0 m (equilibrium), and −0.20 m (turning point).

Three-panel (left-to-right) motion snapshot diagram on a single horizontal baseline track.

Global layout and shared elements (identical in all three panels):
- Draw a straight, perfectly horizontal track as a thick black line spanning nearly the full width of each panel.
- At the far left end of the track in each panel, draw a fixed vertical wall (a thick black rectangle). Label the wall “Fixed wall”.
- A spring is attached to the wall and to the cart. Draw the spring as a zigzag line lying horizontally along the track. Label it “Spring”.
- A cart sits on the track. Draw the cart as a low rectangle with two identical wheels (two circles) touching the track line. Label it “Cart”.
- Above the track, include a position axis for x in each panel: a thin horizontal axis line parallel to the track with an arrow pointing to the right. Label the axis “Position x (m)”.
- On the position axis, show exactly three labeled tick marks: “−0.20”, “0”, and “+0.20”. The “0” tick is centered in the panel. The “−0.20” tick is equally spaced to the left of 0, and the “+0.20” tick is equally spaced to the right of 0.
- Place a small vertical guide line dropped from each tick mark down to the track (faint gray), so the cart can be aligned precisely with the ticks.

Panel 1 (left panel): cart at maximum right displacement and released from rest
- Place the cart so that its center is vertically aligned with the “+0.20” tick mark guide line.
- Draw the spring stretched (longest of the three panels) from the wall to the cart.
- Add a label near the cart: “x = +0.20 m”.
- Add a second label: “released from rest”.
- Draw no velocity arrow (or optionally draw a very short arrow with label “v = 0”, but the text “v = 0” must be explicit if shown).

Panel 2 (middle panel): cart at equilibrium
- Place the cart so that its center is vertically aligned with the “0” tick mark guide line.
- Draw the spring at its natural (intermediate) length compared with the other two panels.
- Add a label near the cart: “x = 0 m (equilibrium)”.

Panel 3 (right panel): cart at maximum left displacement (turning point)
- Place the cart so that its center is vertically aligned with the “−0.20” tick mark guide line.
- Draw the spring compressed (shortest of the three panels) from the wall to the cart.
- Add a label near the cart: “x = −0.20 m”.
- Add a second label: “turning point”.
- Draw no velocity arrow (or explicitly label “v = 0” if a velocity annotation is included).

Consistency constraints:
- The wall position is identical in all three panels.
- The spacing from −0.20 to 0 equals the spacing from 0 to +0.20 on the position axis.
- The cart’s center aligns exactly with the appropriate tick’s guide line in each panel.
- The spring is longest at +0.20 m, intermediate at 0 m, and shortest at −0.20 m.

Figure 2. Energy bar chart at x = +0.20 m (released from rest).

Energy bar chart with two categories, showing kinetic energy K and spring potential energy Uₛ at the instant the cart is released from rest at x = +0.20 m.

Axes:
- X-axis label: “Energy type”.
- X-axis categories: two centered tick labels directly under the bar positions: “K” (left category) and “Uₛ” (right category).
- Y-axis label: “Energy (J)”.
- Y-axis range: from 0 to 2.0 J.
- Y-axis tick marks and labels: 0.0, 0.5, 1.0, 1.5, 2.0 (uniform interval of 0.5 J).
- Origin labeling: the bottom-left corner where the axes meet is labeled “0.0”.
- Axis arrows: arrows on the positive ends of both axes.

Zero-energy reference line:
- Draw a dashed horizontal line exactly at the y-axis tick labeled 0.0 J. This dashed line is the zero-energy line.

Bars (student-completed bars are shown as shaded in the final expected rendering):
- Uniform bar width for both categories: each bar occupies roughly the same width, with a visible gap between the K bar and the Uₛ bar.
- Bar outlines: solid black outline with medium thickness.
- Bar fill: solid medium-gray shading for nonzero energy bars.

Exact bar heights for x = +0.20 m:
- Total mechanical energy for the system is E = 1.6 J.
- K bar (category “K”): height = 0.0 J. Represent this as a distinct solid horizontal line segment exactly on the dashed 0.0 J line at the K category position (no shaded rectangle above zero).
- Uₛ bar (category “Uₛ”): height = 1.6 J, drawn as a shaded rectangle starting exactly at the dashed 0.0 J line and extending upward to the level corresponding to 1.6 J (between the 1.5 J and 2.0 J ticks, closer to 1.5 J).

Error bars (required):
- Draw vertical error bars with caps on top of each category position.
- For the K category: mean = 0.0 J with a nonzero uncertainty; draw an error bar extending upward from 0.0 J to 0.10 J (no portion below 0.0 J). Cap at the top of the error bar. This represents +0.10 J.
- For the Uₛ category: mean = 1.6 J; draw an error bar extending from 1.45 J up to 1.75 J (±0.15 J).
- Error bar style: thin black line.
- Error bar cap width: each cap spans approximately one-third of the bar width.

No grid lines. No title beyond the caption. All numeric tick labels must be visible exactly as listed.

Figure 3. Energy bar chart at x = 0 m (equilibrium).

Energy bar chart with two categories, showing kinetic energy K and spring potential energy Uₛ when the cart is at equilibrium position x = 0 m.

Axes (must match Figure 2 and Figure 4 exactly):
- X-axis label: “Energy type”.
- X-axis categories: “K” (left) and “Uₛ” (right), centered under their respective bar positions.
- Y-axis label: “Energy (J)”.
- Y-axis range: from 0 to 2.0 J.
- Y-axis tick marks and labels: 0.0, 0.5, 1.0, 1.5, 2.0.
- Origin labeled “0.0” at the axes intersection.
- Axis arrows: arrows on the positive ends of both axes.

Zero-energy reference line:
- Dashed horizontal line exactly at 0.0 J.

Bars:
- Uniform bar width for both categories; black outlines; medium-gray fill for nonzero bars.

Exact bar heights for x = 0 m:
- Total mechanical energy for the system is E = 1.6 J.
- Uₛ bar (category “Uₛ”): height = 0.0 J. Represent as a distinct solid horizontal line segment exactly on the dashed 0.0 J line at the Uₛ category position.
- K bar (category “K”): height = 1.6 J. Draw as a shaded rectangle starting exactly at 0.0 J and extending up to 1.6 J.

Error bars (required):
- K category: mean = 1.6 J; error bar extends from 1.45 J to 1.75 J (±0.15 J).
- Uₛ category: mean = 0.0 J; error bar extends upward from 0.0 J to 0.10 J.
- Caps: each cap width is one-third of the bar width.

No grid lines. No extra legends.

Figure 4. Reference energy bar chart at x = −0.10 m (provides the energy scale).

Reference energy bar chart (fully filled in) for the cart–spring system at position x = −0.10 m. This figure defines the exact y-axis scale to be used in Figures 2 and 3.

Axes:
- X-axis label: “Energy type”.
- X-axis categories: “K” (left) and “Uₛ” (right) centered under the bars.
- Y-axis label: “Energy (J)”.
- Y-axis range: 0 to 2.0 J.
- Y-axis tick marks and labels: 0.0, 0.5, 1.0, 1.5, 2.0.
- Origin labeled “0.0”.
- Axis arrows: arrows on the positive ends of both axes.

Zero-energy reference line:
- Dashed horizontal line exactly at 0.0 J.

Bars:
- Uniform bar width; black outlines; solid medium-gray fill.

Exact bar heights at x = −0.10 m (with k = 80 N/m, A = 0.20 m):
- Total mechanical energy E = 1.6 J.
- Spring potential energy Uₛ = 0.40 J. Draw the Uₛ bar to a height exactly at 0.40 J (below the 0.5 J tick, clearly above 0.0).
- Kinetic energy K = 1.20 J. Draw the K bar to a height exactly at 1.20 J (between 1.0 J and 1.5 J ticks, closer to 1.0).

Error bars (required):
- K category: mean = 1.20 J; error bar extends from 1.10 J to 1.30 J (±0.10 J).
- Uₛ category: mean = 0.40 J; error bar extends from 0.35 J to 0.45 J (±0.05 J).
- Cap width: one-third of bar width.

No grid lines. Ensure the K bar is visibly taller than the Uₛ bar, with heights matching the numeric values above.
A.

Draw shaded bars that represent K and UsU_s to complete the energy bar charts in Figure 2 and Figure 3 for when the cart is released from rest at x=+0.20 mx=+0.20\ \text{m} and for when the cart is at x=0 mx=0\ \text{m}, respectively. Figure 4 shows an energy bar chart that represents the kinetic energy K of the cart and the spring potential energy Us of the cart-spring system at the instant that the cart is at x=0.10 mx=-0.10\ \text{m}. The spring potential energy UsU_s is defined to be zero when the cart is at x=0 mx=0\ \text{m}.

• Shaded bars should start at the dashed line that represents zero energy.
• Represent any energy that is equal to zero with a distinct line on the zero-energy line.
• The relative heights of each shaded bar should reflect the magnitude of the respective energy consistent with the scale used in Figure 4.

Figure 5. Cart at x = +0.20 m released from rest, and at x = 0 m moving with maximum speed v_max.

Two-panel diagram (left panel and right panel) showing the same cart–spring system on a horizontal frictionless track, with a clear position reference so x = +0.20 m and x = 0 m are unambiguous.

Shared elements in both panels:
- Draw a horizontal track as a thick black line.
- Draw a fixed wall at the far left end; label it “Fixed wall”.
- Draw a horizontal spring from the wall to the cart; label “Spring”.
- Draw the cart as a rectangle with two wheels; label “Cart”.
- Above the track, draw a position axis labeled “Position x (m)” with a rightward arrow.
- On the position axis, show three labeled ticks with equal spacing: “−0.20”, “0”, “+0.20”. Include faint vertical guide lines from these ticks down to the track.

Left panel (release state):
- Place cart center aligned exactly under the “+0.20” guide line.
- Spring is stretched (longer than in the right panel).
- Text near the cart: “x = +0.20 m” and “released from rest”.
- Velocity annotation: explicitly show “v = 0” (either as text alone or as a zero-length arrow represented by no arrow plus the text “v = 0”).

Right panel (equilibrium, maximum speed):
- Place cart center aligned exactly under the “0” guide line.
- Spring drawn at its natural/intermediate length.
- Text near the cart: “x = 0 m”.
- Draw a horizontal velocity arrow attached to the cart pointing to the left or to the right (either direction is acceptable at equilibrium), and label the arrow exactly “v_max”. The arrow must be clearly nonzero length and strictly horizontal.

No additional numerical values are introduced in this figure beyond the tick labels and x labels above.
B.

Starting with conservation of energy, derive an equation for the maximum speed vmaxv_{\max} of the cart. Express your answer in terms of kk, mm, and the amplitude A=0.20 mA=0.20\ \text{m}. Begin your derivation by writing a fundamental physics principle or an equation from the reference information. Figure 5 shows the cart at x=+0.20 mx=+0.20\ \text{m} when it is released from rest and the cart at x=0 mx=0\ \text{m} when its speed is vmaxv_{\max}. The mass of the cart is 0.50 kg, the spring constant is 80 N/m, and the release position is 0.20 m from equilibrium.

Figure 6. Energy vs. position for the cart–spring system from x = −0.20 m to x = +0.20 m, with Uₛ(x) given; students add E and K(x).

A single Cartesian graph showing energy as a function of position x for a cart–spring oscillator.

Axes (exact formatting):
- Horizontal axis label: “Position x (m)”.
- Horizontal axis range: from −0.20 to +0.20.
- Horizontal tick marks and labels: −0.20, −0.10, 0, +0.10, +0.20 (uniform interval 0.10 m).
- Vertical axis label: “Energy (J)”.
- Vertical axis range: from 0.0 to 2.0.
- Vertical tick marks and labels: 0.0, 0.5, 1.0, 1.5, 2.0 (uniform interval 0.5 J).
- Origin: label “0” at the intersection of the axes (x = 0 and energy = 0.0).
- Axis arrows: arrows on the positive ends of both axes.
- No grid lines.

Given curve (already drawn): spring potential energy Uₛ(x)
- Draw a smooth concave-up U-shaped parabola labeled “Uₛ(x)”.
- The parabola’s minimum touches the x-axis exactly at x = 0 (so Uₛ = 0.0 J at equilibrium).
- The parabola is symmetric about the vertical line through x = 0.
- At x = −0.20 and at x = +0.20, the parabola reaches exactly 1.6 J (the same height on both ends), matching the tick spacing on the y-axis.
- At x = −0.10 and at x = +0.10, the parabola reaches exactly 0.40 J (same height on both sides).

What the student must add (not pre-drawn in the template, but required for the completed figure):
- Total mechanical energy E: a perfectly horizontal straight line labeled “E” at the constant value 1.6 J across the entire x-range from −0.20 to +0.20.
- Kinetic energy K(x): a smooth concave-down arch labeled “K(x)” that equals E − Uₛ(x). This curve:
  - Touches the x-axis (K = 0.0 J) exactly at the two turning points x = −0.20 and x = +0.20.
  - Reaches a single maximum at the center (x = 0) with K = 1.6 J, where it meets the E line.
  - Is symmetric about x = 0.
  - Has K = 1.2 J at x = −0.10 and at x = +0.10.

Line styling:
- Uₛ(x): solid black curve, medium thickness.
- E line: solid black horizontal line, medium thickness, distinct label “E” placed near the right side of the line.
- K(x): solid black curve, medium thickness, label “K(x)” placed near the top of the arch but not overlapping the E label.
C.

Figure 6 shows a graph of the energy of the cart-spring system as a function of the position xx of the cart from x=0.20 mx=-0.20\ \text{m} to x=+0.20 mx=+0.20\ \text{m}. The spring potential energy Us(x)U_s(x) is shown on the graph.

i.

Sketch and label a line that represents the total mechanical energy EE for the cart-spring system as a function of the position xx from x=0.20 mx=-0.20\ \text{m} to x=+0.20 mx=+0.20\ \text{m}.

ii.

Sketch and label a curve that represents the kinetic energy K(x)K(x) for the cart-spring system as a function of the position xx from x=0.20 mx=-0.20\ \text{m} to x=+0.20 mx=+0.20\ \text{m}.

D.

Indicate whether the magnitude of the acceleration a0.10|a_{0.10}| of the cart at x=+0.10 mx=+0.10\ \text{m} is greater than, less than, or equal to the magnitude of the acceleration a0.20|a_{0.20}| at x=+0.20 mx=+0.20\ \text{m}.

a0.10>a0.20|a_{0.10}| > |a_{0.20}|
a0.10<a0.20|a_{0.10}| < |a_{0.20}|
a0.10=a0.20|a_{0.10}| = |a_{0.20}|
Justify how your response is consistent with the energy lines or curves you drew in Figure 6 in part C.

FRQ

Block acceleration in spring-mass systems

4. In Scenario 1, a block of mass m1=0.60 kgm_1 = 0.60\ \text{kg} is attached to a horizontal ideal spring with spring constant k=120 N/mk = 120\ \text{N/m} on a frictionless surface, as shown in Figure 1. The block is pulled to the right to a displacement x=+0.10 mx = +0.10\ \text{m} from equilibrium and released from rest. The block subsequently undergoes simple harmonic motion. All frictional forces are negligible.

In Scenario 2, the spring is attached to a different block of mass m2=0.90 kgm_2 = 0.90\ \text{kg} on the same frictionless surface, as shown in Figure 2. The block is again pulled to the right to the same displacement x=+0.10 mx = +0.10\ \text{m} from equilibrium and released from rest. The block subsequently undergoes simple harmonic motion. All frictional forces are negligible.

Figure 1. Scenario 1: Block–spring system on a frictionless horizontal surface, with the block pulled to x = +0.10 m and released.

Single, static physics apparatus diagram (no motion blur), drawn in clean black lines on a white background.

Overall layout (left to right):
- A vertical wall at the far left edge of the diagram, drawn as a tall rectangle with light diagonal hatch marks to clearly indicate a rigid wall.
- A horizontal surface line extends from the wall to the far right edge of the diagram. The surface is perfectly level.
- Centered above the surface line is a single block attached to the wall by a single ideal spring.

Spring and wall connection:
- The spring is drawn as a horizontal zigzag coil.
- The left end of the spring is firmly attached to the wall at mid-height of the wall (visually centered vertically relative to the block).
- The spring extends rightward from the wall directly to the left face of the block, aligned horizontally (spring axis parallel to the surface).
- A text label placed just above the spring reads exactly: "k = 120 N/m".

Block (Scenario 1 mass):
- A rectangular block sits on the surface line in the right half of the diagram.
- The block’s bottom edge touches the surface line (no gap).
- Centered on the face of the block is the mass label written exactly: "m₁ = 0.60 kg".

Equilibrium position marker (reference position):
- A thin vertical dashed line is drawn on the surface line to mark the equilibrium position of the block’s center (the block’s position when the spring is neither stretched nor compressed).
- This dashed line is placed to the left of the drawn block, indicating the block is currently displaced to the right of equilibrium.
- Immediately below the surface line at this dashed line, add the text label: "equilibrium".

Displacement indication (must show +0.10 m to the right):
- A single horizontal arrow is drawn along the surface line (or just above it), starting exactly at the equilibrium dashed line and ending exactly at the vertical line through the block’s center (the block’s center position as drawn).
- The arrow points to the right, indicating positive displacement.
- Above the arrow, place the text label exactly: "x = +0.10 m".

Direction convention:
- Near the bottom right corner of the diagram, include a small right-pointing arrow labeled exactly "+x" to explicitly define the positive direction as rightward.

Frictionless surface indication:
- Under the surface line, near the center, include the text label exactly: "frictionless surface".

Clarity constraints:
- Only one block is shown (the displaced position). Do NOT show multiple positions.
- The spring is shown stretched relative to equilibrium (consistent with x being positive to the right).

Figure 2. Scenario 2: Same block–spring setup, but with mass m₂ = 0.90 kg; block pulled to the same displacement x = +0.10 m.

Single, static physics apparatus diagram with the SAME geometry, spacing, and labeling conventions as Figure 1 (wall at far left, horizontal surface line, spring attached to wall and to a block on the right). The only intentional change from Figure 1 is the block’s mass label.

Overall layout (left to right):
- A vertical hatched wall at the far left edge.
- A perfectly horizontal surface line extending across the full width.
- A single horizontal zigzag spring attached from the wall to the left face of a single rectangular block resting on the surface line in the right half of the diagram.

Spring label:
- The text label above the spring reads exactly: "k = 120 N/m".

Block (Scenario 2 mass):
- Centered on the block is the mass label written exactly: "m₂ = 0.90 kg".

Equilibrium position marker (reference position):
- A thin vertical dashed line marks the equilibrium position, placed to the left of the drawn block so the block is clearly to the right of equilibrium.
- The dashed line is labeled below the surface line exactly: "equilibrium".

Displacement indication (must match Figure 1 exactly):
- A single horizontal arrow starts exactly at the equilibrium dashed line and ends exactly at the vertical line through the block’s center (block center as drawn).
- The arrow points to the right.
- Above the arrow, the label reads exactly: "x = +0.10 m" (same value and sign as Figure 1).

Direction convention:
- Include the same small right-pointing arrow labeled exactly "+x" near the bottom right.

Frictionless surface indication:
- Under the surface line, include the text label exactly: "frictionless surface".

Clarity constraints:
- Only one block position is shown (the displaced position).
- The spring is shown stretched (consistent with positive rightward displacement).
A.

Refer to Figure 2. Refer to Figure 1. Indicate whether the magnitude of the acceleration of the block at the instant it is released in Scenario 1, a1a_1, is greater than, less than, or equal to the magnitude of the acceleration at the instant it is released in Scenario 2, a2a_2, by writing one of the following in your answer booklet.

a1>a2a_1 > a_2
a1<a2a_1 < a_2
a1=a2a_1 = a_2

Justify your answer in terms of ALL forces exerted on the block at the instant it is released in each scenario. Use qualitative reasoning beyond referencing equations.

B.

Starting with Newton's second law, derive an expression for the instantaneous acceleration aa of the block at the instant it is released. Express your answer in terms of mm, kk, and xx. Begin your derivation by writing a fundamental physics principle or an equation from the reference information. Consider the general case of a block of mass mm attached to a horizontal ideal spring with spring constant kk on a frictionless surface. The block is released from rest from a displacement xx from equilibrium.

C.

Indicate whether the expression for the acceleration aa you derived in part B is or is not consistent with the claim made in part A. Briefly justify your answer by referencing your derivation in part B.

FRQ

Block-spring system oscillation period changes

1. A student investigates the horizontal motion of a block attached to a spring on a frictionless horizontal surface, as shown in Figure 1. The block is attached to the spring, which is fixed to a wall. The block is displaced from equilibrium and released, undergoing simple harmonic motion (SHM).

Figure 1. Block–spring system on a frictionless horizontal surface at the instant of release.

Clean physics apparatus diagram (no perspective), black line art with clear labels.

Overall layout (left to right):
- A vertical wall at the far left edge of the diagram, drawn as a thick vertical rectangle. Label centered on the wall: "Wall".
- A horizontal surface extends rightward from the base of the wall as a single straight thick line across the full width. Place the label "Frictionless surface" below the line near the center.
- A horizontal spring is attached to the wall on its left end and to a block on its right end. The spring is drawn as a coil with uniform spacing between loops.

Reference line and axis:
- Along the surface, draw a horizontal position axis coincident with the surface line (or just above it). Put a right-pointing arrow on this axis labeled "+x" at the arrowhead.
- Mark the equilibrium position with a thin vertical dashed line that intersects the surface. Directly below this dashed line, write "x = 0".

Block and mass label:
- Draw the block as a rectangle resting on the surface line (no wheels). The block’s height is roughly equal to its width (a compact rectangle).
- Put the text "m = 0.500 kg" centered inside the block.

Spring constant label:
- Above the spring, centered along its length, place the text "k = 80.0 N/m".

Initial displacement geometry (must be numerically unambiguous):
- The block is NOT at the equilibrium dashed line; it is displaced to the right.
- Add a second thin vertical dashed line that marks the block’s position at release (the position of the block’s center). This second dashed line is to the right of the equilibrium dashed line.
- Between the two dashed lines (equilibrium and release position), draw a double‑headed horizontal measurement arrow along the surface line.
- Centered above that double‑headed arrow, place the text "+0.120 m".
- Near the release-position dashed line, add the text "t = 0" to indicate the instant shown.

Spring connection clarity:
- The spring’s left end is fixed at the wall.
- The spring’s right end attaches to the left face of the block.
- The spring is drawn stretched compared with its equilibrium length: to enforce this visually, also draw a faint, short spring (or a light gray outline of the spring) corresponding to the equilibrium configuration ending at the equilibrium dashed line; this faint outline is optional ONLY if it does not add new numbers. If included, label it "equilibrium".

Text and labeling rules:
- Only the following numeric values appear anywhere in the figure: "0.500 kg", "80.0 N/m", and "+0.120 m".
- The labels "Wall", "Frictionless surface", "+x", "x = 0", and "t = 0" must be present and clearly associated with their elements.

Figure 2. Axes for a position–time graph x versus t over one full period T.

Blank graph axes only (no data curve), rendered as a standard rectangular plotting area with light grid lines.

Axes and labels (must be explicit and fully numbered):
- Horizontal axis labeled "t (s)" centered below the axis.
- The horizontal axis begins at the left boundary with the tick label "0" and ends at the right boundary with the tick label "T".
- Exactly three equally spaced major tick marks appear between 0 and T, creating four equal time intervals across the axis. The tick labels, from left to right, are: "0", "T/4", "T/2", "3T/4", "T".
- A right-pointing arrowhead is drawn at the positive end of the horizontal axis (just beyond the "T" label).

- Vertical axis labeled "x (m)" rotated vertically along the left axis.
- The vertical axis is symmetric about zero and includes labeled tick marks at five evenly spaced values: "-0.12", "-0.06", "0", "+0.06", "+0.12".
- The origin label "0" is printed at the intersection of the axes (where the horizontal and vertical axes cross).
- An upward-pointing arrowhead is drawn at the positive end of the vertical axis (just above the "+0.12" label).

Grid and styling:
- Light gray grid lines fill the plotting area.
- Major grid lines align with every labeled tick on both axes.
- Minor grid lines (lighter than major) subdivide each major interval into two equal parts on both axes (so each major step is split in half).

No curve requirement:
- No plotted curve, no points, and no markers are present on the axes; the graph is intentionally blank aside from axes, tick marks, tick labels, grid, and axis arrows.

Numeric accuracy constraints:
- The maximum labeled displacement magnitude on the vertical axis is exactly 0.12 m, matching the given initial displacement magnitude in Figure 1.
- The time axis is labeled symbolically in exact fractions of T, not decimals and not seconds values.
A.
i.

On the axes shown in Figure 2, sketch a graph of the position x of the block as a function of time t for one full period from t = 0 to t = T. Be sure to indicate the values of x at t = 0, t = T/4, t = T/2, and t = T.

ii.

Derive an expression for the period T of the oscillations in terms of the mass m and the spring constant k. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

iii.

Derive an expression for the maximum speed v_max of the block in terms of the amplitude A, the mass m, and the spring constant k. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

B.

Indicate whether the period of the oscillations in the modified system increases, decreases, or remains the same compared to the original system. The block-spring system is modified by adding a second spring in parallel with the first spring, as shown in a new setup (not shown). The second spring has spring constant 40.0 N/m. The block is again displaced to x = +0.120 m and released from rest. Assume the springs remain within their elastic limits and the surface remains frictionless.

Increases
Decreases
Remains the same
Justify your response.

Key terms

TermDefinition
displacement from equilibriumThe distance and direction of an object's position relative to its equilibrium position; in SHM, the restoring force is proportional to this quantity.
resonanceThe condition in which a driven oscillator vibrates at maximum amplitude when the driving frequency matches a natural resonant frequency of the system.

Common unit 7 mistakes

Thinking amplitude affects period

Neither T_s = 2*pi*sqrt(m/k) nor T_p = 2*pi*sqrt(L/g) contains amplitude. A larger oscillation takes the same time per cycle as a smaller one. Amplitude does affect total energy and maximum speed, but not period or frequency.

Confusing where velocity and acceleration are maximum

Velocity is maximum at equilibrium (x = 0), not at the turning points. Acceleration is maximum at the turning points (x = plus or minus A), not at equilibrium. These are out of phase by a quarter cycle, and mixing them up leads to wrong answers on graph and scenario questions.

Applying the pendulum period formula outside the small-angle range

T_p = 2*pi*sqrt(L/g) is valid only for small angular displacements where sin(theta) is approximately equal to theta. For large angles, the motion is no longer simple harmonic and this formula does not apply.

Forgetting that pendulum period is independent of bob mass

The formula T_p = 2*pi*sqrt(L/g) contains no mass term. Changing the mass of the pendulum bob does not change the period. Students often assume heavier bobs swing more slowly.

Using total energy as (1/2)*k*x^2 at an arbitrary position

The total energy is (1/2)*k*A^2, where A is the amplitude, not the current displacement x. At an arbitrary position x, the potential energy is (1/2)*k*x^2 and the kinetic energy makes up the rest. Setting total energy equal to (1/2)*k*x^2 at a non-turning-point position is a common algebra error.

How this unit shows up on the AP exam

Predicting the effect of changing system parameters

A common task presents a spring-mass or pendulum system and asks how the period, frequency, or total energy changes when one variable is altered. You need to apply T_s = 2*pi*sqrt(m/k) or T_p = 2*pi*sqrt(L/g) algebraically, explain why amplitude does not appear, and use E_total = (1/2)*k*A^2 to reason about energy changes. Justifying your answer with the relevant equation is expected.

Translating between representations

Multi-part problems often give one representation of SHM, such as an x-t graph or a position equation, and ask you to derive or sketch the corresponding v-t and a-t graphs, identify specific values at labeled points, or describe the motion in words. Knowing the phase relationships between displacement, velocity, and acceleration and being able to identify turning points and equilibrium crossings from any representation is a core skill.

Applying energy conservation to find speed or position

Problems frequently ask for the speed of an oscillating object at a position other than equilibrium or a turning point. The approach is to set (1/2)*k*A^2 equal to (1/2)*m*v^2 + (1/2)*k*x^2 and solve. Variations include changing the amplitude and asking for the new maximum speed, or describing energy bar charts at different points in the cycle and asking for qualitative comparisons.

Final unit 7 review checklist

  • Final Unit 7 review checklistUse this list to confirm you can handle every major skill in the Oscillations unit before exam day.
  • Identify SHM from a force descriptionGiven a force equation or scenario, confirm whether the restoring force is proportional to displacement from equilibrium and directed opposite to it, satisfying m*a_x = -k*delta_x.
  • Calculate period and frequency for both systemsApply T_s = 2*pi*sqrt(m/k) for a spring-mass oscillator and T_p = 2*pi*sqrt(L/g) for a simple pendulum. Predict how changing m, k, L, or g affects the period, and confirm that amplitude changes have no effect.
  • Read and connect SHM graphsGiven an x-t graph, identify where velocity and acceleration are zero, maximum, or minimum. Sketch corresponding v-t and a-t graphs with correct phase relationships and the same period.
  • Apply energy conservation in SHMUse E_total = (1/2)*k*A^2 = (1/2)*m*v^2 + (1/2)*k*x^2 to find speed at any position. Identify where kinetic and potential energies are maximum or zero, and explain how changing amplitude changes total energy.
  • Explain amplitude independence of periodArticulate in words and with equations why the period of a spring-mass system or small-angle pendulum does not depend on amplitude, and why total energy does depend on amplitude.

How to study unit 7

Step 1: Lock in the SHM force conditionStart with Topic 7.1. Read the topic guide and practice identifying whether a described force qualifies as a restoring force. Write out m*a_x = -k*delta_x and explain in your own words why acceleration is largest at the turning points and zero at equilibrium. Apply the same logic to the pendulum case.
Step 2: Practice period and frequency calculationsWork through Topic 7.2 by solving problems that ask you to calculate T or f for spring-mass and pendulum systems, then predict how the period changes when one variable is doubled or halved. Confirm for yourself that amplitude never appears in either period formula.
Step 3: Interpret and sketch SHM graphsFor Topic 7.3, practice sketching x-t, v-t, and a-t graphs for the same oscillator and labeling the turning points, equilibrium crossings, and phase relationships. Given one graph, derive the other two. Use the equations x = A*cos(2*pi*f*t) to check your sketches numerically.
Step 4: Solve energy problems in SHMWork through Topic 7.4 by applying E_total = (1/2)*k*A^2 = (1/2)*m*v^2 + (1/2)*k*x^2 to find speed at intermediate positions. Practice problems where amplitude changes and you must find the new total energy. Connect energy graphs to position graphs by identifying where KE and PE peak.
Step 5: Review with practice questions and FRQ practiceUse the 25+ practice questions and FRQ practice available for this unit to work through multi-part problems that combine force, period, graphs, and energy in a single scenario. Use the AP score calculator to estimate your estimated score range and identify which topic areas need more attention.

More ways to review

Topic study guides

Open the individual guides for Unit 7 when you want a closer review of one topic.

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Practice questions

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FRQ practice

Practice free-response reasoning and compare your answer with scoring guidance.

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Cheatsheets

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Score calculator

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Frequently Asked Questions

What topics are covered in AP Physics 1 Unit 7?

AP Physics 1 Unit 7 covers four topics focused on oscillations and simple harmonic motion: 7.1 Defining Simple Harmonic Motion, 7.2 Frequency and Period of SHM, 7.3 Representing and Analyzing SHM, and 7.4 Energy of Simple Harmonic Oscillators. You'll apply these ideas to spring-object systems and pendulums. The unit connects force, energy, and periodic motion in a way that shows up across many real-world applications. Head to Unit 7 for topic-by-topic practice.

How much of the AP Physics 1 exam is Unit 7?

Unit 7 makes up 5-8% of the AP Physics 1 exam. That weight covers oscillations and simple harmonic motion, including how energy transforms in spring-object systems and pendulums, how to calculate frequency and period, and how to represent SHM graphically and mathematically. It's a smaller unit by topic count (4 topics), but the energy and SHM concepts it introduces connect directly to other units, so a solid understanding pays off on exam day.

What's on the AP Physics 1 Unit 7 progress check (MCQ and FRQ)?

The AP Physics 1 Unit 7 progress check includes both MCQ and FRQ parts drawn from all four unit topics: defining simple harmonic motion, frequency and period, representing and analyzing SHM, and the energy of simple harmonic oscillators. MCQ questions typically ask you to interpret graphs, compare systems, or calculate period and frequency. FRQ questions often ask you to explain energy transformations or justify the motion of a spring or pendulum using SHM principles. Working through the progress check is one of the best ways to spot gaps before the real exam. Find matched practice at Unit 7.

How do I practice AP Physics 1 Unit 7 FRQs?

AP Physics 1 Unit 7 FRQs most often focus on energy in simple harmonic oscillators and representing or analyzing SHM, so those are the two areas to prioritize. Typical question types ask you to sketch or interpret position-time and energy graphs, explain why a restoring force produces SHM, or compare how changing mass or spring constant affects period. To practice effectively, write out full justifications, not just numerical answers. College Board rewards clear reasoning. You can find FRQ-style practice questions at Unit 7.

Where can I find AP Physics 1 Unit 7 practice questions?

The best place to find AP Physics 1 Unit 7 practice questions, including multiple-choice and practice test style problems, is Unit 7. That page organizes MCQ and FRQ practice around all four topics: defining SHM, frequency and period, representing and analyzing SHM, and energy of simple harmonic oscillators. For a practice test experience, work through questions from each topic in one sitting and time yourself. Mixing MCQ and FRQ in the same session mirrors what the real exam feels like.

How should I study AP Physics 1 Unit 7?

Start with the concept of energy in simple harmonic motion, since it ties together nearly every other idea in the unit. Once you understand how kinetic and potential energy trade off in a spring-object oscillator or pendulum, the rest of the unit, including frequency, period, and SHM graphs, clicks into place much faster. Here's a practical study sequence: 1. **Define SHM** (Topic 7.1): Make sure you can explain what a restoring force is and why it produces oscillations. 2. **Frequency and period** (Topic 7.2): Practice deriving and applying the period formulas for springs and pendulums. 3. **Graphs and representations** (Topic 7.3): Sketch position, velocity, and acceleration vs. time graphs from scratch until it feels automatic. 4. **Energy** (Topic 7.4): Work problems where you track energy at different points in the oscillation cycle. After each topic, do a short set of MCQ to check your understanding before moving on. Find topic-aligned practice at Unit 7.

Ready to review Unit 7?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.